Ask question

Ask question

All categories

2 years ago

14

A certain lottery has 38 numbers. In how many different ways can 4 of the numbers be selected? (Assume that order of selection

is not important.)

2 answers:

rodikova [14]2 years ago

8 0

Answer:

There are 73815 ways of selecting 4 of the lottery numbers.

Step-by-step explanation:

The total amount of ways to pick 4 numbers from a set of 38 ignoring the order is given by the combinatorial number of 38 with 4, denoted by ${38 \choose 4}$ , which is

${38 \choose 4} = \frac{38!}{4!(38-4)!} = 73815$

Therefore, there are 73815 ways of picking 4 of the numbers.

diamong [38]2 years ago

8 0

Answer:

We conclude that have 73815 a different ways.

Step-by-step explanation:

We know that a certain lottery has 38 numbers. We assume that order of selection is not important. We calculate how many different ways can 4 of the numbers be selected.

So out of 38 numbers we choose 4 numbers. We get:

$C_4^{38}=\frac{38!}{4!(38-4)!}=73815$

We conclude that have 73815 a different ways.

You might be interested in

Hi! I need help with the attached question in calc. Thank you:)

Elena L [17]

Answer:

3π square units.

Step-by-step explanation:

We can use the disk method.

Since we are revolving around AB, we have a vertical axis of revolution.

So, our representative rectangle will be horizontal.

R₁ is bounded by y = 9x.

So, x = y/9.

Our radius since our axis is AB will be 1 - x or 1 - y/9.

And we are integrating from y = 0 to y = 9.

By the disk method (for a vertical axis of revolution):

$\displaystyle V=\pi \int_a^b [R(y)]^2\, dy$

So:

$\displaystyle V=\pi\int_0^9\Big(1-\frac{y}{9}\Big)^2\, dy$

Simplify:

$\displaystyle V=\pi\int_0^9(1-\frac{2y}{9}+\frac{y^2}{81})\, dy$

Integrate:

$\displaystyle V=\pi\Big[y-\frac{1}{9}y^2+\frac{1}{243}y^3\Big|_0^9\Big]$

Evaluate (I ignored the 0):

$\displaystyle V=\pi[9-\frac{1}{9}(9)^2+\frac{1}{243}(9^3)]=3\pi$

The volume of the solid is 3π square units.

Note:

You can do this without calculus. Notice that R₁ revolved around AB is simply a right cone with radius 1 and height 9. Then by the volume for a cone formula:

$\displaystyle V=\frac{1}{3}\pi(1)^2(9)=3\pi$

We acquire the exact same answer.

8 0

2 years ago

3/4-(-7/6) Write your answer in simplest form.

allsm [11]

Answer:

$\frac{3}{4} - ( - \frac{7}{6} ) \\ = \frac{3}{4} + \frac{7}{6} \\ = \frac{23}{12} \\ = 1 \frac{11}{12}$

6 0

2 years ago

Giving medal. A new car that sells for $21,000 depreciates (decreases in value) 16% each year. Write a function that models the

Natali5045456 [20]

$21,000-16%
16% of $21,000 is $3,360
$21,000-$3,360=$17,640
1st year would be $17,640

16% of $17,640 is $2,822
$17,640-$2,822 is $14,817
2nd year would be $14,817

16% of $14,817 is $2,370
$14,817-$2,370=$12,446
3rd year would be $12,446

$12,446 rounded would be $12,447

So A) $12,447

Hope this helps out :)

4 0

3 years ago

Read 2 more answers

The solution to 4x + 7 - 9 = -14 is a0.

Alchen [17]

Answer:

Isolate the variable by dividing each side by factors that don't contain the variable.

x = − 3

Step-by-step explanation:

4 0

3 years ago

Use mathematical induction to show that 4^n ≡ 3n+1 (mod 9) for all n equal to or greater than 0

cestrela7 [59]

When $n=0$ , you have

$4^0=1\equiv3(0)+1=1\mod9$

Now assume this is true for $n=k$ , i.e.

$4^k\equiv3k+1\mod9$

and under this hypothesis show that it's also true for $n=k+1$ . You have

$4^k\equiv3k+1\mod9$
$4\equiv4\mod9$
$\implies 4\times4^k\equiv4(3k+1)\mod9$
$\implies 4^{k+1}\equiv12k+4\mod9$

In other words, there exists $M$ such that

$4^{k+1}=9M+12k+4$

Rewriting, you have

$4^{k+1}=9M+9k+3k+4$
$4^{k+1}=9(M+k)+3k+3+1$
$4^{k+1}=9(M+k)+3(k+1)+1$

and this is equivalent to $3(k+1)+1$ modulo 9, as desired.

3 0

3 years ago